Problem 93 - Chemistry The Central Science

Question

The vapor pressure of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) at \(19^{\circ} \mathrm{C}\) is 40.0 torr. \(\mathrm{A} 1.00\) -g sample of ethanol is placed in a 2.00 \(\mathrm{L}\) container at \(19^{\circ} \mathrm{C}\) . If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol remain?

Step-by-Step Solution

Answer

At equilibrium, 0.840 g of liquid ethanol remains in the container.

Step 1: Use the Ideal Gas Law to solve for moles of ethanol in vapor phase

Recall that the Ideal Gas Law is given by the equation \[PV = nRT\] where P is the pressure, V is the volume, n is the number of moles of the gas, R is the gas constant, and T is the temperature. We are given the value of P as the vapor pressure of ethanol (40.0 torr) at a temperature of \(19^{\circ} \mathrm{C}\). First, we should convert the pressure into atmospheres to be consistent with the gas constant. \(1 \: \text{atm} = 760 \: \text{torr}\), so, \[P = \frac{40.0 \: \text{torr}}{760 \: \text{torr/atm}} = 0.0526 \: \text{atm}\] Next, we need to convert the temperature to Kelvin: \[T = 19^{\circ} \mathrm{C} + 273.15 = 292.15 \: \text{K}\] Now we have everything we need to solve for the moles of ethanol in the vapor phase, using the Ideal Gas Law. Rearrange the equation to solve for n: \[n = \frac{PV}{RT}\]

Step 2: Plug in the values and solve for n

Now we can plug the given values and constants into the rearranged Ideal Gas Law equation: \[n = \frac{(0.0526 \: \text{atm}) (2.00 \: \text{L})}{(0.0821 \: \text{L atm/mol K})(292.15 \: \text{K})} = 0.00348 \: \text{mol}\]

Step 3: Convert moles of ethanol in the vapor phase to mass

Now that we have calculated the moles of ethanol in the vapor phase (0.00348 mol), we can convert it to mass using the molar mass of ethanol. The molar mass of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), is approximately 46.07 g/mol. Using the conversion factor: \[0.00348 \: \text{mol} \times \frac{46.07 \: \text{g}}{1 \: \text{mol}} = 0.160 \: \text{g}\]

Step 4: Calculate the mass of liquid ethanol remaining

We initially had 1.00 g of ethanol in the container. Now that we know the mass of ethanol in the vapor phase (0.160 g), we can subtract this value from the initial mass to find the mass of liquid ethanol remaining: \[1.00 \: \text{g} - 0.160 \: \text{g} = 0.840 \: \text{g}\] Therefore, 0.840 g of liquid ethanol remains in the container when the system reaches equilibrium.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics used to relate the pressure, volume, temperature, and number of moles of a gas. It is expressed as \[PV = nRT\]where

• P is the pressure of the gas
• V is the volume occupied by the gas
• n is the number of moles of gas
• R is the ideal gas constant (0.0821 \( ext{L atm/mol K} \)
• T is the temperature in Kelvin

This equation helps us understand how gases behave under various conditions by providing a mathematical relationship between these quantities. To ensure consistent units, pressure is often converted into atmospheres and temperature into Kelvin before calculations are made.
For instance, when working with the vapor pressure of ethanol, which was given in torr, converting it into atmospheres was necessary for applying the Ideal Gas Law effectively.

Moles of Gas

The concept of moles is central to chemistry as it allows us to quantify the amount of a substance. One mole corresponds to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities, whether atoms, molecules, or ions. In the context of gases and particularly liquid-vapor interactions, the moles represent the quantity of gas in the vapor phase.
In the problem, we used the Ideal Gas Law to find the number of moles of ethanol in the vapor phase. This was achieved by rearranging the equation to \[n = \frac{PV}{RT}\]allowing the calculation when pressure, volume, and temperature are known. Calculating moles in this way provides a link between measurable physical properties and chemical quantities, bridging practical experimentation with theory.

Equilibrium

Equilibrium in chemistry refers to a situation where the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the system. In a closed container with ethanol, equilibrium is reached when the vapor and liquid phases have balanced each other out.
At equilibrium, the amount of ethanol evaporating into the vapor equals the amount re-condensing back into the liquid. This dynamic balance means that macroscopically, the composition of each phase remains constant.
The calculation of remaining liquid ethanol involved understanding that only a specific amount of ethanol could exist as vapor under given conditions, dictated by the equilibrium vapor pressure. Hence, equilibrium is crucial for predicting the final distribution of substance in a closed system.

Vapor-Liquid Equilibrium

Vapor-liquid equilibrium is a state where a liquid and its vapor coexist at a given temperature and pressure, without further changes in their amounts. This concept is pivotal in understanding phase transitions and properties like vapor pressure.
When ethanol is placed in a closed container, it will evaporate until reaching vapor-liquid equilibrium. At this point, the vapor pressure of ethanol is exactly exerted by a sufficient number of molecules in the gas phase balancing those in the liquid phase.
This concept explains why after some ethanol vaporizes, not all liquid transforms into vapor. The limitations are set by the molecules' ability to exert a certain pressure, known as vapor pressure, which is unique to each substance at a given temperature. Understanding vapor-liquid equilibrium helps predict how much of each phase will be present under equilibrium conditions, vital for calculations like those in the problem statement.